$A=2x^2+x+5$
$=2(x^2+\frac{x}{2}+\frac{5}{2})$
$=2(x^2+2.\frac{1}{4}.x+\frac{1}{16})+\frac{39}{8}$
$=2(x+\frac{1}{4})^2+\frac{39}{8}≥\frac{39}{8}∀x$
Dấu ''='' xảy ra khi $x+\frac{1}{4}=0⇔x=-\frac{1}{4}$
Vậy $A_{min}=\frac{39}{8}⇔x=-\frac{1}{4}$
$B=-3x^2+x+1$
$=-3(x^2-\frac{x}{3}-\frac{1}{3})$
$=-3(x^2-2.\frac{1}{6}.x+\frac{1}{36})+\frac{13}{12}$
$=-3(x-\frac{1}{6})^2+\frac{13}{12}≤\frac{13}{12}∀x$
Dấu ''='' xảy ra khi $x-\frac{1}{6}=0⇔x=\frac{1}{6}$
Vậy $B_{max}=\frac{13}{12}⇔x=\frac{1}{6}$.
